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จงหาค่าของ \( \sum_{i=1}^{30}(i+4)(i-3) \)

Ask by Hobbs Romero. in Thailand
Feb 25,2025

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ค่าของ \( \sum_{i=1}^{30}(i+4)(i-3) \) คือ 9,560

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Calculate or simplify the expression \( \sum_{i=1}^{30}(i+4)(i-3) \). Find the sum by following steps: - step0: Solution: \(\sum _{i=1}^{30}\left(i+4\right)\left(i-3\right)\) - step1: Convert the expressions: \(\left(1+4\right)\left(1-3\right)+\left(2+4\right)\left(2-3\right)+\left(3+4\right)\left(3-3\right)+\left(4+4\right)\left(4-3\right)+\left(5+4\right)\left(5-3\right)+\left(6+4\right)\left(6-3\right)+\left(7+4\right)\left(7-3\right)+\left(8+4\right)\left(8-3\right)+\left(9+4\right)\left(9-3\right)+\left(10+4\right)\left(10-3\right)+\left(11+4\right)\left(11-3\right)+\left(12+4\right)\left(12-3\right)+\left(13+4\right)\left(13-3\right)+\left(14+4\right)\left(14-3\right)+\left(15+4\right)\left(15-3\right)+\left(16+4\right)\left(16-3\right)+\left(17+4\right)\left(17-3\right)+\left(18+4\right)\left(18-3\right)+\left(19+4\right)\left(19-3\right)+\left(20+4\right)\left(20-3\right)+\left(21+4\right)\left(21-3\right)+\left(22+4\right)\left(22-3\right)+\left(23+4\right)\left(23-3\right)+\left(24+4\right)\left(24-3\right)+\left(25+4\right)\left(25-3\right)+\left(26+4\right)\left(26-3\right)+\left(27+4\right)\left(27-3\right)+\left(28+4\right)\left(28-3\right)+\left(29+4\right)\left(29-3\right)+\left(30+4\right)\left(30-3\right)\) - step2: Calculate: \(-10-6+0+8+18+30+44+60+78+98+120+144+170+198+228+260+294+330+368+408+450+494+540+588+638+690+744+800+858+918\) - step3: Calculate: \(9560\) ค่าของ \( \sum_{i=1}^{30}(i+4)(i-3) \) คือ 9560

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เรามาเริ่มกันที่การขยายผลของ \((i+4)(i-3)\) ก่อนนะครับ \[ (i+4)(i-3) = i^2 - 3i + 4i - 12 = i^2 + i - 12 \] จากนั้นเราสามารถแยกผลรวมออกเป็น 3 ส่วนได้ดังนี้: \[ \sum_{i=1}^{30}(i+4)(i-3) = \sum_{i=1}^{30} (i^2 + i - 12) = \sum_{i=1}^{30} i^2 + \sum_{i=1}^{30} i - \sum_{i=1}^{30} 12 \] ในการคำนวณ: 1. **ผลรวมของ \(i^2\):** \[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \] สำหรับ \(n=30\): \[ \sum_{i=1}^{30} i^2 = \frac{30(30+1)(2 \cdot 30+1)}{6} = \frac{30 \cdot 31 \cdot 61}{6} = 9455 \] 2. **ผลรวมของ \(i\):** \[ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \] สำหรับ \(n=30\): \[ \sum_{i=1}^{30} i = \frac{30(30+1)}{2} = \frac{30 \cdot 31}{2} = 465 \] 3. **ผลรวมของ \(12\):** \[ \sum_{i=1}^{30} 12 = 12 \cdot 30 = 360 \] ตอนนี้เราสามารถนำผลมารวมเข้าด้วยกัน: \[ \sum_{i=1}^{30}(i+4)(i-3) = 9455 + 465 - 360 \] คำนวณผลรวมทั้งหมด: \[ = 9455 + 465 - 360 = 10380 \] สุดท้าย ดังนั้นค่าของ \( \sum_{i=1}^{30}(i+4)(i-3) = 10380 \) ครับ!

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